Graphs of Groups and Bass Serre Theory - Examples

View $A\ast B$ as the graph of groups with two vertices, vertex groups $A$ and $B$, and one trivial edge group between them. Then [L1] gives vertex cosets $(A\ast B)\mathrm{/}A$ and $(A\ast B)\mathrm{/}B$, while the edge cosets are just the elements of $A\ast B$ itself.

By [L2], the quotient is a single segment. Thus each edge joins one $A$-coset to one $B$-coset, giving the usual bipartite Bass-Serre tree of the free product.

Realize $A{\ast }_{C}B$ by the one-segment graph of groups from [L1]. Then [L2] says the two vertex orbits are $(A{\ast }_{C}B)\mathrm{/}A$ and $(A{\ast }_{C}B)\mathrm{/}B$, while the edge orbit is $(A{\ast }_{C}B)\mathrm{/}C$.

The edge coset $\gamma C$ joins the vertices $\gamma A$ and $\gamma B$, so the tree records exactly how the amalgamating subgroup sits in the two factors.

Because $m,n\mathrm{\ne }0$, the maps $\mathbb{Z}\to \mathbb{Z}$, $k\mapsto mk$ and $k\mapsto nk$, are injective. Applying [L1] to these two monomorphisms gives exactly the presentation of $BS(m,n)$.

Therefore [L2] describes its Bass-Serre tree: vertices are left cosets of the vertex copy of $\mathbb{Z}$, edges are left cosets of the edge copy of $\mathbb{Z}$, and the stable letter translates between adjacent levels of this tree.

Translation by one step on the bi-infinite line has one vertex orbit and one edge orbit, so by [L2] the quotient graph is a single loop. The action is free and without inversions.

Therefore [L1] identifies the acting group with a free group on one generator, namely the loop edge of the quotient graph. This recovers $\mathbb{Z}$.

The subgroup $H=C_2$ already is one of the free-product factors of $G=C_2\ast C_3$. Therefore one of the Kurosh intersection terms in [L1] is exactly $H$.

No further nontrivial intersection factor is needed, and there is no free remainder. So the Kurosh decomposition here is the tautological one $H\cong C_2$.

By [L1], the fundamental group $\mathrm{\Gamma }=C_2\ast C_3$ acts on its Bass-Serre tree with vertex stabilizers conjugate to $C_2$ and $C_3$ and trivial edge stabilizers. The canonical surjection $\mathrm{\Gamma }\to C_2\times C_3$ has finite image of order $6$, so its kernel ${\mathrm{\Gamma }}_0$ has index $6$. Because this quotient map is injective on each factor, ${\mathrm{\Gamma }}_0$ meets every conjugate of $C_2$ and $C_3$ trivially.

The subgroup ${\mathrm{\Gamma }}_0$ therefore acts freely on the same tree, so [L2] makes ${\mathrm{\Gamma }}_0$ a free group. Since ${\mathrm{\Gamma }}_0$ has finite index in $\mathrm{\Gamma }$, the group $C_2\ast C_3$ is virtually free.

Take the same one-loop quotient graph twice. In one graph of groups put the trivial vertex group and trivial edge group; the resulting group is $\mathbb{Z}$. In the other put vertex group $C_2$ and trivial edge group; the resulting group is $C_2\ast \mathbb{Z}$.

The quotient graphs are identical but the resulting groups are not isomorphic, so the quotient graph alone does not determine the acting group.